Bulletin of TICMI
Vol. 17, No. 1, 2013, 19–27
The Solutions of the Boundary Value Problems of the Theory of
Thermoelasticity with Microtemperatures for an Elastic Circle
Ivane Tsagarelia∗
a
I.Vekua Institute of Applied Mathematics
of Iv. Javakhishvili Tbilisi State University,
University Str. 2, Tbilisi 0128, Georgia
(Received December 4, 2012; Revised January 30, 2013; Accepted March 5, 2013)
In the present work, using absolutely and uniformly convergent series, the 2D boundary value
problems of statics of the linear theory of thermoelasticity with microtemperatures for an
elastic circle are solved explicitly. The question on the uniqueness of a solution of the problem
is investigated.
Keywords: Thermoelasticity, Microtemperatura, Boundary value problem, Explicit
solution.
AMS Subject Classification: 74F05, 74G30, 74G10.
1.
Introduction
Together with generalization and development along several paths, the linear theory of thermoelasticity with microtemperatures has recently attracted considerable
effort directed toward mathematical research and construction of explicit solutions
for boundary value problems in specific domains. Of the publications devoted to
such problems, we note [1,2], which also contain historical and bibliographic information.
2.
Basic equations and boundary value problem
Consider a circle D of radius R with boundary S. Find a regular vector U =
(u1 , u2 , u3 , w1 , w2 ), (U ∈ C 1 (D) ∩ C 2 (D), D = D ∪ S) satisfying in the circle D a
system of equations [1,2]:
µ∆u(x) + (λ + µ)graddivu(x) = βgradu3 (x),
k∆u3 (x) + k1 divw(x) = 0,
k6 ∆w(x) + (k4 + k5 )graddivw(x) − k3 gradu3 (x) − k2 w(x) = 0,
and on the circumference S one of the following conditions:
∗ Corresponding
author. Email: i.tsagareli@yahoo.com
ISSN: 1512-0082 print
c 2013 Tbilisi University Press
⃝
(1)
Bulletin of TICMI
20
I. u(z) = f (z), u3 (z) = f3 (z), T ′′ (∂z , n)w(z) = p(z);
II. u(z) = f (z), k
∂u3 (z)
+ k1 w(z)n(z) = f3 (z), T ′′ (∂z , n)w(z) = p(z);
∂n(z)
(2)
III. T ′ (∂z , n)u(z) − βu3 (z)n(z) = f (z), u3 (z) = f3 (z), T ′′ (∂z , n)w(z) = p(z),
where u(x) is the displacement vector of the point x, u = (u1 , u2 ); w = (w1 , w2 ) is
the microtemperatures vector; u3 is temperature measured from the constant absolute temperature T0 ; n is the external unit normal vector to S; f = (f1 , f2 ), p =
(p1 , p2 ), f1 , f2 , f3 -are the given functions on S; λ, µ, β, k, k1 , k2 , k3 , k4 , k5 , k6 are
constitutive coefficients [1,2]; T ′ u is the stress vector in the classical theory of
elasticity; T ′′ w is stress vector for microtemperatures [2]:
∑
∂u(x)
ni (x)gradui (x),
+ λn(x)divu(x) + µ
∂n
i=1
(3)
2
∑
∂w(x)
′′
T (∂x , n)w(x) = (k5 + k6 )
ni (x)gradwi (x).
+ k4 n(x)divw(x) + k5
∂n
2
T ′ (∂x , n)u(x) = µ
i=1
Separately we will study the following problems:
1. Find in a circle D solution u(x) of equation (1)1 , if on the circumference S
there are given the values:
of the vector u
(problem A1 );
of the vector T ′ (∂x , n)u(x) − βu3 (x)n(x)
(problem A2 ).
2. Find in the circle D solutions u3 (x) and w(x) of the system of equations (1)2
and (1)3 , if on the circumference S there are given the values:
of the function u3 (z) and the vectorT ′′ (∂z , n)w(z) (problem K1 );
∂u3 (z)
+ k1 w(z)n(z) and the vector T ′′ (∂z , n)w(z) (problem K2 ).
of the function k
∂n(z)
Thus the above-formulated problems of thermoelasticity with microtemperatures
can be considered as a union of two problems: I- (A1 , K1 ), II- (A1 , K2 ) and III(A2 , K1 ).
3.
Uniqueness theorems
Let (u′ , u′3 , w′ ) and (u′′ , u′′3 , w′′ ) be two different solutions of any of the problems
I, II, III. Then the differences u = u′ − u′′ , u3 = u′3 − u′′3 and w = w′ − w′′ of
these solutions, obviously, satisfy the homogeneous system (1)0 and zero boundary
conditions (2)0 . For a regular solutions of equation (1)1 and equations (1)2 and
(1)3 the Green’s formulas [2,3]:
Vol. 17, No. 1, 2013
∫
∫
u(y)[T ′ (∂y , n)u(y) − βu3 (y)n(y)]dy S,
[E1 (u(x), u(x)) − βu3 (x)divu(x)]dx =
D
21
S
∫
[T0 E2 (w(x), w(x)) + k | gradu3 |2 +(k1 + k3 T0 )wgradu3 + k2 T0 | w(x) |2 ]dx =
D
∫
u3 (y)[k
S
∂u3 (y)
+ k1 w(y)n(y)] + T0 w(y)[T ′′ (∂y , n)w(y)]dy S,
∂n(y)
(4)
is valid [2], where
( ∂u
( ∂u
∂u2 )2
∂u2 )2
∂u2 )2
1
1
+µ
+µ
;
−
+
∂x1 ∂x2
∂x1 ∂x2
∂x2 ∂x1
( ∂w
( ∂w
∂w2 )2
∂w2 )2
1
1
E2 (w, w) = 12 (2k4 + k5 + k6 )
+
+ (k6 + k5 )
−
∂x1
∂x2
∂x1
∂x2
( ∂w
)2
( ∂w
)2
∂w
∂w
1
2
2
1
+(k6 + k5 )
+ (k6 − k5 )
,
+
−
∂x2
∂x1
∂x1
∂x2
E1 (u, u) = (λ + µ)
( ∂u
1
+
under the conditions that: λ + µ, µ > 0, 2k4 + k5 + k6 > 0, k6 ± k5 > 0, E1 and E2
are nonnegative quadratic forms [3].
Taking into account formula (4)2 and the homogeneous boundary conditions for
the problems Ki , (i = 1, 2), we obtain E2 (w, w) = 0, gradu3 = 0, u3 = 0, w = 0.
The solution of the above equations has the form: u3 (x) = const, w = 0.
The following theorems are valid.
Theorem 3.1 : The difference of two arbitrary solutions of problem K1 is equal
to zero: w(x) = 0, u3 (x) = 0, x ∈ D.
The difference of two arbitrary solutions of problem K2 may differ only by an
arbitrary constant: w(x) = 0, u3 (x) = const, x ∈ D.
Taking into account Theorem 3.1 and formula (4)1 , under the homogeneous
boundary conditions for the problems I, II and III we obtain E1 (u, u)−βu3 divu =
0. The solution of the above equation, when u3 = 0 or u3 = const, has the form
u1 (x) = −c1 x2 + q1 ,
u2 (x) = c1 x1 + q2 ,
(5)
where c1 , q1 and q2 are arbitrary constants.
The following theorems are valid.
Theorem 3.2 : The difference of two arbitrary solutions of problem I is the vector
U (u1 (x), u2 (x), u3 (x), w1 (x), w2 (x)), where u1 = u2 = 0, u3 = 0, w1 = w2 = 0.
Theorem 3.3 : The difference of two arbitrary solutions of problem II is the
vector U (u1 (x), u2 (x), u3 (x), w1 (x), w2 (x)), where u1 = u2 = 0, u3 = c, w1 =
w2 = 0; c is an arbitrary constant.
Theorem 3.4 : The difference of two arbitrary solutions of problem III is the
vector U (u1 (x), u2 (x), u3 (x), w1 (x), w2 (x)), where u1 and u2 are expressed by formulas (5), and u3 = 0, w1 = w2 = 0.
Bulletin of TICMI
22
4.
Solutions of the Problems
On the basis of the system [(1)2 , (1)3 ], we can write
△(△ + s21 )u3 = 0,
△(△ + s21 )divw = 0.
Solutions of these equations are represented in the form [4]:
u3 (x) = φ1 (x) + φ2 (x),
w1 (x) = a1
∂φ1 (x)
∂φ2 (x)
∂φ3 (x)
+ a2
− a3
,
∂x1
∂x2
∂x2
w2 (x) = a1
∂φ1 (x)
∂φ2 (x)
∂φ3 (x)
+ a2
+ a3
,
∂x2
∂x1
∂x1
where △φ1 = 0, (△ + s21 )φ2 = 0, (△ + s22 )φ3 = 0, s21 = −
s22 = −
k2
k3
k
k6
, a1 = − , a 2 = − , a 3 = ;
k6
k2
k1
k7
kk2 − k1 k3
,
kk7
k7 = k4 +k5 +k6 ;
Problem K1 . Taking into account formulas:
(6)
k, k2 , k6 , k7 > 0 [2].
∂
n1 ∂
∂
= n2
+
,
∂x2
∂r
r ∂ψ
∂
=
∂x1
n2 ∂
∂
−
, we rewrite the representations (6) and the boundary conditions
∂r
r ∂ψ
of the problemK1 in the tangent and normal components:
n1
u3 (x) = φ1 (x) + φ2 (x),
wn (x) = a1
∂
∂
1 ∂
φ1 (x) + a2 φ2 (x) − a3
φ3 (x),
∂r
∂r
r ∂ψ
ws (x) = a1
1 ∂
1 ∂
∂
φ1 (x) + a2
φ2 (x) + a3 φ3 (x);
r ∂ψ
r ∂ψ
∂r
[
φ1 (z) + φ2 (z) = f3 (z), k7
[
k6
∂ws
∂r
]
[
+
R
k5 ∂wn
R ∂ψ
]
∂wn
∂r
]
+
R
= ps (z),
(7)
[
]
k4 ∂ws
= pn (z),
R ∂ψ R
(8)
R
where wn = (w ·n), ws = (w ·s), pn = (p·n), ps = (p·s), n = (n1 , n2 ), s = (−n2 , n1 ),
∂
∂
=
.
∂n
∂r
The harmonic function φ1 and metaharmonic functions φ2 and φ3 are represented
Vol. 17, No. 1, 2013
23
in the form of series in the circle [5]:
∞ ( )
∑
r m
1
(Ym1 · νm (ψ)),
φ1 (x) = Y01 +
2
R
m=1
∞
∑
φ2 (x) = I0 (s2 r)Y02 +
Im (s2 r)(Ym2 · νm (ψ)),
(9)
m=1
∞
∑
φ3 (x) = I0 (s3 r)Y03 +
Im (s3 r)(Ym3 · sm (ψ)),
m=1
respectively, where Ymk are the unknown two-component constants vectors,
νm (ψ) = (cos mψ, sin mψ), sm (ψ) = (− sin mψ, cos mψ), k = 1, 2, m = 0, 1, ....
Let the functions pn , ps and f3 expand into the Fourier series:
∞
α0 ∑
+
(αm · νm (ψ)),
2
m=1
∞
γ0 ∑
(γm · νm (ψ)),
f3 (z) =
+
2
pn (z) =
ps (z) =
∞
β0 ∑
+
(βm · sm (ψ)),
2
m=1
(10)
m=1
where
αm = (αm1 , αm2 ), βm = (βm1 , βm2 ), γm = (γm1 , γm2 ),
αm1
∫2π
1
=
π
pn (θ) cos(mθ)dθ,
0
∫2π
αm2
∫2π
1
1
=
pn (θ) sin(mθ)dθ, βm1 =
ps (θ) cos(mθ)dθ,
π
π
0
0
∫2π
βm2 =
1
ps (θ) sin(mθ)dθ,
π
0
γm1
1
=
π
∫2π
f3 (θ) cos(mθ)dθ, γm2
0
1
=
π
∫2π
f3 (θ) sin(mθ)dθ.
0
We substitute (9) into (7) and then the obtained expression and (10) into (8).
Passing to the limit, as r → R, for the unknowns Ymk we obtain a system of
algebraic equations:
1
γ0
α0
β0
Y01 + I0 (s2 R)Y02 = , k7 a2 s22 I0′′ (s2 R)Y02 =
, k6 a3 s23 I0′′ (s3 R)Y03 = ; (11)
2
2
2
2
Bulletin of TICMI
24
Ym1 + Im (s2 R)Ym2 = γm ,
′′
a1 m[(k7 − k4 )m − k7 ]Ym1 + a2 [k7 s22 Im
(s2 R)R2 − k4 m2 a1 Im (s2 R)]Ym2
′
′
+a3 m[k7 [s3 RIm
(s3 R) − Im (s3 R)] − k4 Rs3 Im
(s3 R)]Ym3 = αm R2 ,
a1 m[(k6 + k5 )m − k6 ]Ym1
(12)
′
′
+a2 m[k6 [s2 RIm
(s2 R) − Im (s2 R)] + k5 Rs2 Im
(s2 R)]Ym2
′′
+a3 [k6 R2 s23 Im
(s3 R) + k5 a3 m2 Im (s3 R)] = βm R2 ,
m = 1, 2, ....
Relying on the theorem on the uniqueness of a solution of the problem we can
conclude that the principal determinants of systems (11) and (12) are other than
zero. Substituting the solutions of systems (11) and (12) into (9) and then into (6),
we can find values of the functions u3 (x), w1 (x) and w2 (x).
Problem K2 . Taking into account formulas (6), the boundary conditions of the
problem K2 can be rewritten as:
[
k
∂u3
∂r
[
k6
]
∂ws
∂r
[
+ k1 [wn ]R = f3 (z),
k7
R
]
[
+
R
k5 ∂wn
R ∂ψ
∂wn
∂r
]
+
R
[
]
k4 ∂ws
= pn (z),
R ∂ψ R
(13)
]
= ps (z).
R
We substitute (9) into (7), then the obtained expression and (10) into (13).
Passing to the limit, as r → R, from (13) we obtain the system of linear algebraic
equations with regard to the unknowns Ymk for every value m:
k7 a2 s22 I0′′ (s2 R)Y02 =
α0
,
2
s2 I0′ (s2 R)(k + k1 a2 )Y02
k6 a3 s23 I0′′ (s3 R)Y03 =
β0
,
2
(14)
γ0
= ;
2
′
m(k + k1 a1 )Ym1 + s2 Im
(s2 R)(k + k1 a2 R)Ym2 + a3 mIm (s3 R)Ym3 = γm R,
′′
a1 m[(k7 − k4 )m − k7 ]Ym1 + a2 [k7 s22 Im
(s2 R)R2 − k4 m2 a1 Im (s2 R)]Ym2
′
′
+a3 m[k7 [s3 RIm
(s3 R) − Im (s3 R)] − k4 Rs3 Im
(s3 R)]Ym3 = αm R2 ,
a1 m[(k6 + k5 )m − k6 ]Ym1
(15)
′
′
+a2 m[k6 [s2 RIm
(s2 R) − Im (s2 R)] + k5 Rs2 Im
(s2 R)]Ym2
′′
+a3 [k6 R2 s23 Im
(s3 R) + k5 a3 m2 Im (s3 R)] = βm R2 , m =
1, 2, ....
From equation (1)2 , taking into account the boundary conditions (2) and formulae (10) we can write
∫
∫
[k∆u3 (x) + k1 divw(x)]dx =
D
[k
S
∂u3 (y)
+ k1 w(y)n(y)]dy S = 0,
∂n(y)
Vol. 17, No. 1, 2013
γ01
1
=
π
25
∫2π
f3 (θ)dθ = 0.
0
For the Y02 we obtain:
α01
1
=
π
Y02 = 0; then
∫2π
pn (θ)dθ = 0,
β01
0
1
=
π
∫2π
ps (θ)dθ = 0;
Y03 = 0,
Y01 = const.
0
Problem A1 . A solution (1)1 is sought in the form
u(x) = v0 (x) + v(x),
(16)
where v0 is a particular solution of equation (1)1 , and v is a general solution of the
corresponding homogeneous equation (1)1 . Direct checking shows that v0 has the
form
v0 (x) =
β
1
grad[− 2 φ2 (x) + φ0 (x)],
λ + 2µ
s1
(17)
where φ0 is a biharmonic function: △φ0 = φ1 .
A solution v(x) = (v1 (x), v2 (x)) of the homogeneous equation corresponding to
(1)1 :
µ△v(x) + (λ + µ)graddivv(x) = 0
is sought in the form
∂
[Φ1 (x) + Φ2 (x)] −
∂x1
∂
v2 (x) =
[Φ1 (x) + Φ2 (x)] +
∂x2
v1 (x) =
∂
Φ3 (x),
∂x2
∂
Φ3 (x),
∂x1
(18)
where
∆Φ1 (x) = 0,
∆∆Φ2 (x) = 0,
∆∆Φ3 (x) = 0,
(λ + 2µ)
∂
∂
∆Φ2 (x) − µ
∆Φ3 (x) = 0,
∂x1
∂x2
(λ + 2µ)
∂
∂
∆Φ2 (x) + µ
∆Φ3 (x) = 0,
∂x2
∂x1
(19)
Φ1 , Φ2 , Φ3 are the scalar functions.
Taking into account (16) condition (2)1 , we can write
v(z) = Ψ(z),
(20)
where Ψ(z) = f (z) − v0 (z) is the known vector, v0 is defined by formula (17), and
φ1 and φ2 by equalities (9), where the value of the Ymk vectors is defined by means
Bulletin of TICMI
26
of systems (11) and (12). The function φ0 is a solution of the equation △φ0 = φ1 ;
it has the form
∞
R2 ∑ 1 ( r )m+2
(Ym1 · νm (ψ)),
φ0 (x) =
4
m+1 R
(21)
m=0
where Ym1 are defined from (11) and (12).
In view of (19), we can represent the harmonic function Φ1 and biharmonic
functions Φ2 and Φ3 in the form
∞ ( )
∑
∞ ( r )m
∑
r m+2
(Xm1 · νm (ψ)), Φ2 (x) =
(Xm2 · νm (ψ)),
R
m=0 R
m=0
(22)
∞
R2 (λ + 2µ) ∑ ( r )m+2
Φ3 (x) =
(Xm2 · sm (ψ)),
µ
R
Φ1 (x) =
m=0
where Xmk are the unknown two-component vectors, k = 1, 2.
Substituting (22) into (18), the obtained expressions into (20), we obtain the
system of algebraic equations for every m, whose solution is written as follows:
ς0 R
ηm R (ςm − ηm )R
η0 R
, X02 =
, Xm1 =
−
,
4
4(λ + 2µ)
m
2(λ + µ)m
(ςm − ηm )R
=µ
, m = 1, 2, . . . ;
2(λ + µ)m
X01 =
Xm2
(23)
where ηm and ςm are the Fourier coefficients of the function Ψ(z):
ηm = (ηm1 , ηm2 ),
ηm1
1
=
π
ςm1 =
1
π
η0 = (η01 , 0),
ςm = (ςm1 , ςm2 ),
∫2π
Ψn (θ) cos mθdθ,
γm2
1
=
π
0
∫2π
Ψs (θ) cos mθdθ,
0
δm2 =
1
π
ς0 = (ς01 , 0),
∫2π
Ψs (θ) sin mθdθ,
(24)
0
∫2π
Ψn (θ) sin mθdθ;
0
Ψn and Ψs are normal and tangential components of the function Ψ(z), respectively.
Thus the solution of problem A1 is represented by the sum (16) in which v(x) is
defined by means of formula (18), and v0 (x) by formula (17).
Problem A2 . Taking into account (16) condition (2)3 , we can rewrite it as
T ′ (∂z , n)v(z) = Ψ(z),
(25)
where Ψ(z) = f (z)+βu3 (z)n(z)−T ′ (∂z , n)v0 (z) is the known vector, Ψ = (Ψ1 , Ψ2 ).
We substitute (22) first into (18) and then into (25). For the unknowns Xm1 and
Xm2 we obtain a system of algebraic equations whose solution has the form
Vol. 17, No. 1, 2013
27
η0 R 2
ς0 R 2
R2
c4 R2
, X02 =
, Xm1 =
ςm −
(µηm − c1 ςm ),
4(λ + 2µ)
4(λ + 2µ)
c3
c2 c3 − c1 c4
c4 R 2
Xm2 =
(µηm − c1 ςm ),
c2 c3 − c1 c4
where c1 = µ[2(λ + µ)m2 − (λ + 2µ)m], c2 = 2(λ + µ)(λ + 3µ)m2 + (λ + 2µ)[(3λ +
5µ)m+2µ], c3 = mµ(2µ−1), c4 = 2(λ+3µ)m(2m+3)+2(λ+2µ), m = 1, 2, ....
ηm and ςm are the Fourier coefficients of respectively normal and tangential components of the function Ψ(z).
Having solved problems A1 , A2 , K1 and K2 , we can write solutions of the initial
problems I, II and III.
X01 =
References
[1] D. Iesan and R. Quintanilla, On the theory of thermoelasticity with microtemperatures, J. Thermal
Stresses, 23 (2000), 199-215
[2] A. Scalia, M. Svanadze, and R. Tracina. Basic theorems in the equilibrium theory of thernoelasticity
with microtemperatures, J. Thermal Stresses, 33 (2010), 721-753
[3] N. Muskhelishvili, Some Basic Problems of the Mathematical Theory of Elasticity, Noordhoff, Groningen, Holland, 1953.
[4] I.N. Vekua, On metaharmonic functions, Tr. Tbilis. Mat. Inst., 12 (1943), 105-174 (Russian).
[5] I.N. Vekua, A New Methods of Solution for Elliptic Equations, Fizmatgiz, Moscow-Leningrad, 1948
(Russian).